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Math Problems
Algebra 2
Simplify expressions using trigonometric identities
Simplify to a single trig function with no denominator.
\newline
5
sec
2
x
−
5
tan
2
x
5\sec^{2}x-5\tan^{2}x
5
sec
2
x
−
5
tan
2
x
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\newline
What is the particular solution to the differential equation
d
y
d
x
=
1
+
y
x
\frac{d y}{d x}=\frac{1+y}{x}
d
x
d
y
=
x
1
+
y
with the initial condition
y
(
e
)
=
1
y(e)=1
y
(
e
)
=
1
?
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Solve for the angle
θ
\theta
θ
.
\newline
(a)
sin
2
θ
=
3
4
\sin^{2}\theta=\frac{3}{4}
sin
2
θ
=
4
3
.
\newline
(b)
sin
2
θ
−
cos
θ
=
0
\sin 2\theta-\cos \theta=0
sin
2
θ
−
cos
θ
=
0
.
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
5
x
2
y
4
−
4
x
2
y
3
=
3
−
2
x
-5 x^{2} y^{4}-4 x^{2} y^{3}=3-2 x
−
5
x
2
y
4
−
4
x
2
y
3
=
3
−
2
x
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Solve the equation for all real solutions in simplest form.
\newline
−
3
r
2
−
r
−
10
=
−
4
r
2
-3 r^{2}-r-10=-4 r^{2}
−
3
r
2
−
r
−
10
=
−
4
r
2
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1
−
cos
α
sin
α
=
2
csc
α
\frac{1-\cos \alpha}{\sin \alpha}=2 \csc \alpha
s
i
n
α
1
−
c
o
s
α
=
2
csc
α
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
7
cos
(
3
x
)
sin
(
3
y
)
=
4
x
−
1
-7 \cos (3 x) \sin (3 y)=4 x-1
−
7
cos
(
3
x
)
sin
(
3
y
)
=
4
x
−
1
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
−
y
−
7
y
3
−
4
x
4
+
6
x
−
2
x
y
=
1
-y-7 y^{3}-4 x^{4}+6 x-2 x y=1
−
y
−
7
y
3
−
4
x
4
+
6
x
−
2
x
y
=
1
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Simplify to a single trig function with no denominator.
\newline
csc
θ
sin
θ
\frac{\csc \theta}{\sin \theta}
sin
θ
csc
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
csc
θ
sec
θ
\frac{\csc \theta}{\sec \theta}
sec
θ
csc
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
θ
tan
θ
\frac{\sin \theta}{\tan \theta}
tan
θ
sin
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cot
2
θ
csc
2
θ
\frac{\cot ^{2} \theta}{\csc ^{2} \theta}
csc
2
θ
cot
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
2
θ
⋅
csc
θ
\sin ^{2} \theta \cdot \csc \theta
sin
2
θ
⋅
csc
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cot
θ
csc
θ
\frac{\cot \theta}{\csc \theta}
csc
θ
cot
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cot
θ
cos
θ
\frac{\cot \theta}{\cos \theta}
cos
θ
cot
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
2
θ
⋅
cot
2
θ
\sin ^{2} \theta \cdot \cot ^{2} \theta
sin
2
θ
⋅
cot
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cos
θ
⋅
sec
θ
\cos \theta \cdot \sec \theta
cos
θ
⋅
sec
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
sin
2
θ
⋅
sec
2
θ
\sin ^{2} \theta \cdot \sec ^{2} \theta
sin
2
θ
⋅
sec
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cos
θ
sec
θ
\frac{\cos \theta}{\sec \theta}
sec
θ
cos
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
tan
2
θ
sec
2
θ
\frac{\tan ^{2} \theta}{\sec ^{2} \theta}
sec
2
θ
tan
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
tan
2
θ
⋅
csc
2
θ
\tan ^{2} \theta \cdot \csc ^{2} \theta
tan
2
θ
⋅
csc
2
θ
\newline
Answer:
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Simplify to a single trig function with no denominator.
\newline
cos
θ
cot
θ
\frac{\cos \theta}{\cot \theta}
cot
θ
cos
θ
\newline
Answer:
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Solve for the exact value of
x
x
x
.
\newline
4
ln
(
6
x
−
4
)
+
7
=
15
4 \ln (6 x-4)+7=15
4
ln
(
6
x
−
4
)
+
7
=
15
\newline
Answer:
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Simplify the expression
\newline
tan
(
π
2
−
θ
)
sin
θ
\tan \left(\frac{\pi}{2}-\theta\right) \sin \theta
tan
(
2
π
−
θ
)
sin
θ
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Find the exact value of each of the remaining trigonometric functions of
θ
\theta
θ
.
tan
θ
=
−
1
5
,
sec
θ
<
0
\tan \theta = -\frac{1}{5},\quad \sec \theta < 0
tan
θ
=
−
5
1
,
sec
θ
<
0
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Using implicit differentiation, find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
\newline
−
7
x
2
y
4
−
5
x
y
2
=
4
−
4
x
-7x^{2}y^{4}-5xy^{2}=4-4x
−
7
x
2
y
4
−
5
x
y
2
=
4
−
4
x
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Select the answer which is equivalent to the given expression using your calculator.
\newline
cot
(
arccos
315
18
)
\cot \left(\arccos \frac{\sqrt{315}}{18}\right)
cot
(
arccos
18
315
)
\newline
3
315
\frac{3}{\sqrt{315}}
315
3
\newline
315
18
\frac{\sqrt{315}}{18}
18
315
\newline
315
3
\frac{\sqrt{315}}{3}
3
315
\newline
18
3
\frac{18}{3}
3
18
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Select the answer which is equivalent to the given expression using your calculator.
\newline
csc
(
arccos
15
4
)
\csc \left(\arccos \frac{\sqrt{15}}{4}\right)
csc
(
arccos
4
15
)
\newline
4
15
\frac{4}{\sqrt{15}}
15
4
\newline
4
4
4
\newline
15
4
\frac{\sqrt{15}}{4}
4
15
\newline
15
\sqrt{15}
15
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Select the answer which is equivalent to the given expression using your calculator.
\newline
sin
(
arcsin
18
424
)
\sin \left(\arcsin \frac{18}{\sqrt{424}}\right)
sin
(
arcsin
424
18
)
\newline
18
424
\frac{18}{\sqrt{424}}
424
18
\newline
10
18
\frac{10}{18}
18
10
\newline
424
18
\frac{\sqrt{424}}{18}
18
424
\newline
18
10
\frac{18}{10}
10
18
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For the following sequence determine the common difference (if it is an arithmetic sequence) or the common ratio (if it is a geometric sequence).
\newline
12
,
8
3
,
16
,
…
12, \quad 8 \sqrt{3}, \quad 16, \ldots
12
,
8
3
,
16
,
…
\newline
2
3
2 \sqrt{3}
2
3
\newline
3
3
\frac{\sqrt{3}}{3}
3
3
\newline
3
\sqrt{3}
3
\newline
2
3
3
\frac{2 \sqrt{3}}{3}
3
2
3
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Let
y
y
y
be defined implicitly by the equation
\newline
(
−
8
x
−
2
y
)
3
=
−
x
2
−
10
y
2
.
(-8 x-2 y)^{3}=-x^{2}-10 y^{2} .
(
−
8
x
−
2
y
)
3
=
−
x
2
−
10
y
2
.
\newline
Use implicit differentiation to find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
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Using implicit differentiation, find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
(
3
x
3
−
3
y
2
)
2
=
5
x
y
\left(3 x^{3}-3 y^{2}\right)^{2}=5 x y
(
3
x
3
−
3
y
2
)
2
=
5
x
y
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If
θ
=
4
π
9
\theta=\frac{4\pi}{9}
θ
=
9
4
π
radians, what is the value of
θ
\theta
θ
in degrees?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
0
∘
20^{\circ}
2
0
∘
\newline
(B)
3
6
∘
36^{\circ}
3
6
∘
\newline
(C)
8
0
∘
80^{\circ}
8
0
∘
\newline
(D)
72
0
∘
720^{\circ}
72
0
∘
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A curve is defined by the parametric equations
x
(
t
)
=
t
2
+
9
t
x(t)=t^{2}+9 t
x
(
t
)
=
t
2
+
9
t
and
y
(
t
)
=
−
3
t
2
−
10
t
+
4
y(t)=-3 t^{2}-10 t+4
y
(
t
)
=
−
3
t
2
−
10
t
+
4
. Find
d
y
d
x
\frac{d y}{d x}
d
x
d
y
.
\newline
Answer:
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The graph of
y
=
−
g
(
x
)
y=-g(x)
y
=
−
g
(
x
)
in the
x
y
x y
x
y
-plane always has a negative slope and passes through the origin. If
g
g
g
is an exponential function, which of the following could define
g
g
g
?
\newline
Choose
1
1
1
answer:
\newline
(A)
g
(
x
)
=
−
(
2
)
x
+
1
g(x)=-(2)^{x}+1
g
(
x
)
=
−
(
2
)
x
+
1
\newline
(B)
g
(
x
)
=
−
(
3
4
)
x
+
1
g(x)=-\left(\frac{3}{4}\right)^{x}+1
g
(
x
)
=
−
(
4
3
)
x
+
1
\newline
(C)
g
(
x
)
=
(
2
5
)
x
−
1
g(x)=\left(\frac{2}{5}\right)^{x}-1
g
(
x
)
=
(
5
2
)
x
−
1
\newline
(D)
g
(
x
)
=
3
x
g(x)=3^{x}
g
(
x
)
=
3
x
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Can this differential equation be solved using separation of variables?
\newline
d
y
d
x
=
(
7
−
3
x
)
e
−
y
\frac{d y}{d x}=(7-3 x) e^{-y}
d
x
d
y
=
(
7
−
3
x
)
e
−
y
\newline
Choose
1
1
1
answer:
\newline
(A) Yes
\newline
(B) No
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Can this differential equation be solved using separation of variables?
\newline
d
y
d
x
=
sin
(
y
)
+
y
4
x
+
7
\frac{d y}{d x}=\frac{\sin (y)+y}{4 x+7}
d
x
d
y
=
4
x
+
7
sin
(
y
)
+
y
\newline
Choose
1
1
1
answer:
\newline
(A) Yes
\newline
(B) No
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Can this differential equation be solved using separation of variables?
\newline
d
y
d
x
=
x
y
\frac{d y}{d x}=\sqrt{x y}
d
x
d
y
=
x
y
\newline
Choose
1
1
1
answer:
\newline
(A) Yes
\newline
(B) No
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Can this differential equation be solved using separation of variables?
\newline
d
y
d
x
=
3
x
y
+
5
y
\frac{d y}{d x}=\sqrt{3 x y+5 y}
d
x
d
y
=
3
x
y
+
5
y
\newline
Choose
1
1
1
answer:
\newline
(A) Yes
\newline
(B) No
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Can this differential equation be solved using separation of variables?
\newline
d
y
d
x
=
2
x
+
1
3
y
\frac{d y}{d x}=\frac{2 x+1}{3 y}
d
x
d
y
=
3
y
2
x
+
1
\newline
Choose
1
1
1
answer:
\newline
(A) Yes
\newline
(B) No
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The differentiable functions
x
x
x
and
y
y
y
are related by the following equation:
\newline
1
y
=
8
−
x
\frac{1}{y}=8-x
y
1
=
8
−
x
\newline
Also,
d
y
d
t
=
−
0.5
\frac{d y}{d t}=-0.5
d
t
d
y
=
−
0.5
.
\newline
Find
d
x
d
t
\frac{d x}{d t}
d
t
d
x
when
y
=
0.2
y=0.2
y
=
0.2
.
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The differentiable functions
x
x
x
and
y
y
y
are related by the following equation:
\newline
1
y
=
cos
(
x
)
\frac{1}{y}=\cos (x)
y
1
=
cos
(
x
)
\newline
Also,
d
x
d
t
=
−
2
\frac{d x}{d t}=-2
d
t
d
x
=
−
2
.
\newline
Find
d
y
d
t
\frac{d y}{d t}
d
t
d
y
when
x
=
π
x=\pi
x
=
π
.
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Let
g
(
x
)
=
sin
(
x
)
g(x)=\sin (x)
g
(
x
)
=
sin
(
x
)
.
\newline
Can we use the mean value theorem to say the equation
g
′
(
x
)
=
2
π
g^{\prime}(x)=\frac{2}{\pi}
g
′
(
x
)
=
π
2
has a solution where
0
<
x
<
π
2
0<x<\frac{\pi}{2}
0
<
x
<
2
π
?
\newline
Choose
1
1
1
answer:
\newline
(A) No, since the function is not differentiable on that interval.
\newline
(B) No, since the average rate of change of
g
g
g
over the interval
0
≤
x
≤
π
2
0 \leq x \leq \frac{\pi}{2}
0
≤
x
≤
2
π
isn't equal to
2
π
\frac{2}{\pi}
π
2
.
\newline
(C) Yes, both conditions for using the mean value theorem have been met.
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